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Question Number 167221 by rexford last updated on 09/Mar/22

T h e p l a n e y = 1 s l i c e s t h e s u r f a c e

z = a r c t a n (\frac{x + y}{1 - x y})

i n a c u r v e C .

F i n d t h e s l o p e o f t h e t a n g e n t l i n e t o

C a t x = 2

Answered by TheSupreme last updated on 10/Mar/22

z = a r c t a n (\frac{x + 1}{1 - x})

z (2) = a r c t a n (- 3)

z^{'} = - \frac{2}{(1 + {(\frac{1 + x}{1 - x})}^{2}) {(1 - x)}^{2}}

z = z^{'} (2) (x - 2) + z (2)

z = - \frac{1}{5} (x - 2) + a r c t a n (- 3)

Commented by rexford last updated on 10/Mar/22

t h a n k y o u

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